Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
const int base = 15;
vector<int> adj[maxn];
vector<int> gr[maxn][20];
int anc[maxn][20];
vector<int> gr1[maxn];
int cnt[maxn];
int dep[maxn];
vector<int> mrg(vector<int> nw1, vector<int> nw2) {
  int id1 = 0;
  int id2 = 0;
  vector<int> ans;
  int pre = -1e15;
  while (1) {
    while (id1 < nw1.size() && nw1[id1] <= pre) id1++;
    while (id2 < nw2.size() && nw2[id2] <= pre) id2++;
    if (id1 < nw1.size() && id2 < nw2.size() && ans.size() < base) {
      if (nw1[id1] < nw2[id2]) {
        ans.push_back(nw1[id1]);
        pre = ans.back();
        id1++;
      } else {
        ans.push_back(nw2[id2]);
        pre = ans.back();
        id2++;
      }
    } else {
      break;
    }
  }
  while (1) {
    while (id1 < nw1.size() && nw1[id1] <= pre) id1++;
    if (ans.size() < base && id1 < nw1.size()) {
      ans.push_back(nw1[id1]);
      pre = ans.back();
      id1++;
    } else {
      break;
    }
  }
  while (1) {
    while (id2 < nw2.size() && nw2[id2] <= pre) id2++;
    if (ans.size() < base && id2 < nw2.size()) {
      ans.push_back(nw2[id2]);
      pre = ans.back();
      id2++;
    } else {
      break;
    }
  }
  return ans;
}
void dfs(int u, int par) {
  anc[u][0] = par;
  gr[u][0].clear();
  gr[u][0] = mrg(gr1[par], gr[u][0]);
  for (int i = 1; i <= 18; i++) {
    gr[u][i] = mrg(gr[u][i - 1], gr[anc[u][i - 1]][i - 1]);
    anc[u][i] = anc[anc[u][i - 1]][i - 1];
  }
  cnt[u] += (int)(gr1[u].size());
  for (auto to : adj[u]) {
    if (to == par) continue;
    cnt[to] = cnt[u];
    dep[to] = dep[u] + 1;
    dfs(to, u);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  int kc = dep[x] - dep[y];
  for (int i = 18; i >= 0; i--) {
    if (kc & (1ll << i)) {
      x = anc[x][i];
    }
  }
  if (x == y) return x;
  for (int i = 18; i >= 0; i--) {
    if (anc[x][i] != anc[y][i]) {
      x = anc[x][i];
      y = anc[y][i];
    }
  }
  return anc[x][0];
}
int top_e(int l, int y) { int p = dep[y] - dep[l] - 1; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (fopen("atlarge.in", "r")) {
    freopen("atlarge.in", "r", stdin);
    freopen("atlarge.out", "w", stdout);
  }
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    gr1[x].push_back(i);
  }
  gr1[0].clear();
  for (int i = 1; i <= n; i++) {
    sort(gr1[i].begin(), gr1[i].end());
    gr1[i].erase(unique(gr1[i].begin(), gr1[i].end()), gr1[i].end());
    while (gr1[i].size() > base) {
      gr1[i].pop_back();
    }
  }
  dfs(1, 0);
  while (q--) {
    int x, y, a;
    cin >> x >> y >> a;
    if (dep[x] < dep[y]) swap(x, y);
    int l = lca(x, y);
    if (y == l) {
      vector<int> ans;
      int kc = dep[x] - dep[l];
      ans = mrg(ans, gr1[x]);
      for (int i = 18; i >= 0; i--) {
        if (kc & (1ll << i)) {
          ans = mrg(ans, gr[x][i]);
          x = anc[x][i];
        }
      }
      int k = min((int)ans.size(), a);
      cout << k << " ";
      for (int i = 1; i <= k; i++) {
        cout << ans[i - 1] << " ";
      }
      cout << "\n";
    } else {
      vector<int> ans;
      ans.clear();
      int kc = dep[x] - dep[l];
      ans = mrg(ans, gr1[x]);
      ans = mrg(ans, gr1[y]);
      for (int i = 18; i >= 0; i--) {
        if (kc & (1ll << i)) {
          ans = mrg(ans, gr[x][i]);
          x = anc[x][i];
        }
      }
      kc = dep[y] - dep[l] - 1;
      for (int i = 18; i >= 0; i--) {
        if (kc & (1ll << i)) {
          ans = mrg(ans, gr[y][i]);
          y = anc[y][i];
        }
      }
      int k = min((int)ans.size(), a);
      cout << k << " ";
      for (int i = 1; i <= k; i++) {
        cout << ans[i - 1] << " ";
      }
      cout << "\n";
    }
  }
}